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Standard Cell Potentials

>> Parts of this equation/concept include:

A - Calculating E°cell
B - Relationship to G°
C - Relationship to K

 

A. Calculating E°cell

The standard cell potential (E°cell) is the difference between the reduction potentials of the two half–reactions.

E°cell = E°c – E°a         (Equation 17.11)

The reduction potential of the oxidation reaction is E°a (a for "anode") and the reduction potential of the reduction reaction is E°c (c for "cathode"). Using this equation, the numbers are used directly from the table of standard reduction potentials without modification for multiplication to match the number of electrons or changing sign for reversing reactions. In fact, the change of sign is accomplished with the equation, so a separate step would undo that change. The only hard part of using this equation is resisting the temptation to make it harder.

>> Example 1

What is the standard cell potential of the following cells?

  1. O2 + 4 H+ + 4 e– 2 H2O

    H2CO + H2O HCO2H + 2 H+ + 2 e–

  2. Sn0 Sn2+ + 2 e–

    2 e– + 2 H+ + H2O2 2 H2O

  3. Ni + 2 OH– Ni(OH)2 + 2 e–

    SO42– + 4 H2O + 6 e– S + 8 OH–

Solution:

  1. The reduction of oxygen is the cathode reaction. Its reduction potential is +1.2291 V. The oxidation of carbon is the anode reaction. Its reduction potential is +0.237 V. So

    E°cell = E°c – E°a = +1.2291 – +0.237 = +0.922 V

    Recall that the significant–figure rule for subtraction is that the answer has the fewest number of decimal places.

  2. The oxidation of tin is the anodic reaction with a reduction potential of –0.141 V. The reduction of oxygen is the cathodic reaction with a reduction potential of +1.763 V. So

    E°cell = E°c – E°a = +1.763 – (–0.141) = + 1.904 V

  3. The oxidation of nickel is the anodic reaction with a reduction potential of –0.714 V. The reduction of sulfur is the cathodic reaction with a reduction potential of –0.751 V. So

    E°cell = E°c – E°a = –0.751 – (–0.741) = – 0.010 V

 

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B. Relationship to G°

The relationship between G and E is

G = –nFE         (Equation 17.6)

where n = number of electrons in the total reaction, F is Faraday's constant (9.65 x 104 C/mol), and E is the cell potential. Under standard conditions,

G° = –nFE°

>> Example 2

What is the free energy of the following cells under standard conditions?

  1. O2 + 4 H+ + 4 e– 2 H2O

    H2CO + H2O HCO2H + 2 H+ + 2 e–

  2. Sn0 Sn2+ + 2 e–

    2 e– + 2 H+ + H2O2 2 H2O

  3. Ni + 2 OH– Ni(OH)2 + 2 e–

    SO42– + 4 H2O + 6 e– S + 8 OH–

Solution:

The standard cell potentials were determined in Example 1.

  1. Before the two half–reactions are combined, each must be multiplied by some factor so that the number of electrons is the same in each half. The least common multiple of electrons for these reactions is 4. In Example 1, E° = +0.922 V. Therefore for the reaction

    O2 + 2 H2CO 2 HCO2H

    The standard free energy G° is

    G° = –nFE° = –(4 mol e–)(9.65 x 104 C/mol e–)(0.922 V) = –3.56 x 105 C•V

    Since 1 C•V = 1 J

    G° = –3.56 x 105 J

  2. When the two reactions are combined, and without canceling the electrons,

    2 e– + 2 H+ + H2O2 + Sn Sn2+ + 2 H2O + 2 e–

    Using Equation 17.6,

    G° = –nFE° = –(2)(9.65 x 104)(+1.904) = –3.67 x 105 J

  3. For the overall reaction

    6 e– + 3 Ni + 4 H2O + SO42– 3 Ni(OH)2 + S + 2 OH– + 6 e–

    Using Equation 17.6,

    G° = –nFE = –(6)(9.65 x 104)(–0.010) = +5.8 x 103 J

 

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C. Relationship to K

As G° is r elated to the equilibrium constant (K), so is E°. As with G°, it is the standard state only that is related to the equilibrium constant. Equation 17.19 assumes that the temperature is 298 K.

log K  = 
nE°cell
0.0592
         (Equation 17.19)

>> Example 3

What is the equilibrium constant of the following reactions?

  1. O2 + 2 H2CO 2 HCO2H
  2. 2 H+ + H2O2 + Sn Sn2+ + 2 H2O
  3. 3 Ni + 4 H2O + SO42– 3 Ni(OH)2 + S + 2 OH–

Solution:

The E°cell were determined in Example 1. The number of electrons transferred was determined in Example 2. The value of 0.0592 was determined from F/RT when T = 298 K. Therefore this value has three significant figures. Remember that the number of decimal places in a log term becomes the number of significant figures in the antilog.

  1. log K  = 
    nE°cell
    0.0592
    		 = 
    (4)(0.922)
    0.0592
    		 =  62.297  (to the appropriate significant figures = 62.3)

    K = 1062.3 = 2 x 1062

  2. log K  = 
    nE°cell
    0.0592
    		 = 
    (2)(1.904)
    0.0592
    		 =  64.324  (to the appropriate significant figures = 64.3)

    K = 1064.3 = 2 x 1064

  3. log K  = 
    nE°cell
    0.0592
    		 = 
    (6)(–0.010)
    0.0592
    		 =  –1.0135  (to the appropriate significant figures = –1.0)

    K = 10–1.0 = 0.1

 

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